Did you know that climate science and mainstream academia believes that the Sun can’t heat the Earth? It is true. The field of climate physics and mainstream science believes that sunshine can only heat any location on the Earth to -18 degrees Celsius or 0F, and that any temperatures on the Earth higher than that are from the air warming itself up some more. Physics teaches today that you can cook a frozen dinner at four-times the power but for one-quarter the time listed on the directions, or conversely, at one-quarter the power but four-times the cooking time. Apparently, the culinary arts has something to teach mainstream physics!
Climate of Sophistry 
Joe, thanks for taking the effort to explain the basics to a person, Roy Spencer in this case, who is so deeply stuck in his own world that he will never see the reality. We are doomed by the stupiditiy of the intelligencia. You deserve a Nobel Prize for Physics!
Yes, now write a book titled, Basic Reality For PhDs, um, … maybe too advanced, though, … possibly a better idea would be to write a children’s book for them, with a silly title, with big print and silly cartoon characters in there somehow.
Do a children’s show for PhDs, where you dress as a character that they can happily identify with, … a clown, of course:
Hi, profs! [really high-pitched voice] Today let’s talk about sunshine! [Keep the sentences short and your energy of delivery always high and clown-like] The sun is our friend. It keeps us warm. It makes our weather. Plants drink its life-giving rays. All these wonderful things happen because the sun is strong, not weak. It has the power to make stuff right now. You can’t slice up this power, because it happens right now. You can’t slice up right now, because it’s right now, in this moment. See the beautiful sun, with all its power, making these wonderful things. [Here, maybe have a goofy sidekick come in, to keep the kiddies from getting bored]
“We are doomed by the stupidity of the intelligentsia.” – indeed! We will fix it.
I love the children’s show for PhD’s! They absolutely deserve something like that. In fact, I think it is the only thing that would work…the condescension would be lost on them, and they would show it to their grad students for course material…lol.
“You can’t slice up right now, because it’s right now, in this moment.” – lost it…lol
Joe as a southern boy I am an expert on the cooking of animals. I’ve looked 27 slabs of ribs at one time on 3 55gal charcoal homemade grills. 375 to 400 for about 2 hour give or take bbq sauce and turns will make you a rock star. Had I tried to cook the same meal at even 175 I would be ridiculed instead of praised. At that temperature meat would never cook through. Go grow potatoes
That’s makin’ me hungry!
And yes, exactly! Why not cook it at 90F but start a week earlier? Climate physics says you can do it that way…
From the comments on the vid:
Adam Birnbaum
1 hour ago
Why are you pretending to be a scientist?
Joseph Postma
1 second ago
Please cook your meat at 90F but start a week earlier. Let us know how it turns out…from the hospital.
Science has no method for self-correction other than for generations to die (according to quantum physicist Max Planck)…
Postma, Joseph. In the Cold Light of Day: Flat Earth in Modern Physics and a Numerical Proof for God: A Climate Alarm Story . Kindle Edition.
Reminds you of something ?
You must really be throttled on YTube two people and one is a troll? Yeah something weird about that.
I commented at the YouTube site [I have a different handle over there], and the comment posted four times in a row, when I just submitted one time. But I guess what I said bears repeating — even the universe knows this, I suppose. (^_^)
“Do a children’s show for PhDs” here’s one I wrote in December 2015: http://www.ilovemycarbondioxide.com/Klaas-Vaak/pdf/Klaas-Vaak.pdf
Hans S,
Here’s what Klaas has to say:
Hah, excellent, thanks Robert.
Hi Joe Excellent post as usual I was just wounder if you have seen John Christy’s latest at GWPF https://www.thegwpf.com/putting-climate-change-claims-to-the-test/ Same old flat earth same old
“Will the IPCC AR6 investigate this long-running mismatch between climate systems and models, and the climate system of the real world? I don’t make predictions – hardly at all – but here’s something I might think about.
Past responses have been:
1) The observations are wrong, the models are right, it’s the observations that are wrong.
2) The models had bad forcing; if they had the right forcing then they would get the observations.
But never 3) the models are failed hypotheses.
I predict 3 will not be the choice they choose, although the scientific methods that I grew up with and perhaps you grew up with, say that’s exactly the conclusion you are supposed to make when you do a test like this.”
LOL – the models are failed hypotheses – ya think!!!!!???? Hahaha
Hi, Just found this dandy…
https://www.open.edu/openlearn/nature-environment/climate-change/content-section-1.2.1
Usually they just tell you that the Sun’s flux is divided by 4 without much explanation. Here they not only tell you why but how. The Sun hits the disk just as strongly at the poles as at the equator. I wonder how they can explain why the pole are cold and the equator is hot ?
Could someone analyze
https://sites.google.com/site/floriankapsenberg/matlab-projects/solar-heat-flux-by-latitude
There’s a matlab script to play with. I don’t have that software.
I’m not sure that I understand your question correctly but anyway. If you look at Rosco,s figure 7 (I know Rosco, it’s not yours ) 🙂 You see that no matter the tilt of the Earth, the half sphere will always receive the same amount of energy. Since the earth is tilted at a 23,5 degrees, at the equinox it will be the equator that receives max energy. At the winter solstice (northern hemisphere), it will be the -23,5 degree lattitude (Capricorne) that will rreceive max energy. At the summer solstice. it will be the +23,5 degree lattitude (Cancer) that will receive max energy. This explains not only why the equator is hot and the poles are cold but also the seasons. Hope it was the answer you were looking for.
Regards
From that link from Pierre:
“However, the Earth is spherical , so the area presented to the incoming solar radiation by the rotating Earth (over a period of 24 hours or more) is 4πR 2; i.e. four times as great. Thus, the solar input per unit area averaged over the surface area of the whole Earth is a quarter of the solar constant;”
You see the internal contradiction here? They’re talking about the solar constant which is *ALWAYS* listed in units of Joules per second per square meter (W/m^2), i.e. the *per-second input*, but then they refer that numerical value to 24 hours.
So no…the “area presented to the incoming solar radiation by the rotating Earth” WHEN USING THE PER-SECOND INPUT is a hemisphere, with a projection incidence angle varying as the cosine of the zenith angle.
These people are ***king brain-dead…I mean seriously, they are literally retarded.
Joules per second per square meter is a RATE OF FLOW isn’t it?
How can you have a rate where there is no flow?!
I have three, five-gallon buckets sitting on a one-acre plot of land. I place a hose in each bucket, running at a rate of one gallon per minute per five-gallon bucket. Can I now say that the average rate of flow over the acre of land is some dilution of this RATE for the buckets?
I spray a hose on a window of the front side of a house, at a rate of one gallon per minute. Can I say that I am washing the windows of the whole house with an average of water over all the windows of the house that the water NEVER touches?
There is no rate on the dark side of the Earth. The rate happens where there is light. The effects of the rate can influence the dark side, but the rate itself does not exist there in any amount whatsoever.
Looking at the link Zoe provided I downloaded this simple image :-
Note it shows the solar insolation as a series of vectors normal to the tangential plane at various latitudes – the green arrows – over the hemisphere.
“We begin the process by finding the normal components of solar heat flux for each latitude and longitude. If the normal component is negative, it means that the sun is below the horizon, and it is therefore night; therefore the negative components can be ignored as there can be no negative solar heat flux on the planet’s surface. For each latitude the daily average heat flux is the average heat flux for all longitudes. As the earth moves around the sun, the normal component changes for each latitude, and therefore so does the daily average heat flux.”
Also note that as December means the Sun is effectively at 23.5°S – a few degrees from my home – and December corresponds to Perihelion in Earth’s orbit the solar “constant” is ~1408 W/m2 TOA.
The heading says “Average Annual Solar Heat Flux (Planet Axial Tilt = 23.5°) and it shows a maximum of 400 W/m2. I do not know what mathematical transformations were performed to calculate these average values (I don’t have the software either) but they do not reflect real time insolation. Note “Average” !
All you need to do is to consider a solar panel rated at 15.6% efficiency, a surface area of ~1.2 m2 and a 200 Watt rating.
This panel requires ~1068 W/m2 insolation to justify its 200 W rating and mine actually state this explicitly – “Values at Standard Test Conditions STC (Air Mass AM1.5, irradiance 1000W/m², Cell Temperature 25°C).”
Reduce insolation by 4 and it rates as a 64 Watt panel.
Who’s not telling the truth about real time solar insolation – the Solar panel industry or climate science ?
“Joseph,
I’m curious, so what is the average temp @ 23.5 tilt? It’s hard for to eyeball it from his diagram. Too lazy to do the math myself”
I’m too lazy to do it full math also, but forgetting higher flux at Perihelion, accelleration and decellaration to and from the Perihelion, Etc, Etc, Etc…
At Equinox the Capricorne Tropic is at -23,5 angle from the Sun. 1368*Cos(23.5) = 1254.
At Summer Solstice the Capricorne Tropic is at 0 angle from the Sun, 1368 * Cos(0) = 1368.
At Winter Solstice the Capricorne Tropic is at -47 angle from the sun, 1368 * Cos(47) = 933.
Since the Capricorne Tropic passes just as much time above as below the average angle of -23,5 in a symmetrical Sine wave fashion, I guestimate… (1254 * 2 + 1368 + 933) /4 = 1202 W/m^2.
Probably a bit lower since 933 is further down 1254 then 1254 is from 1368 but since the flux is higher at Perihelion !?!?!?.
Just a approximate approximation.
This is an abstract from a paper I meant to read but each time I read the abstract I just lose interest in the rest. The magic gas strikes again.
“CO2 is the strongest anthropogenic forcing agent for climate change since pre-industrial times. Like other greenhouse gases, CO2 absorbs terrestrial surface radiation and causes emission from the atmosphere to space. As the surface is generally warmer than the atmosphere, the total long-wave emission to space is commonly less than the surface emission. However,this does not hold true for the high elevated areas of central Antarctica. For this region, it is shown that the greenhouse effect of CO2 is around zero or even negative. Moreover, for central Antarctica an increase in CO2 concentration leads to an increased long-wave energy loss to space, which cools the earth-atmosphere system. These unique findings for central Antarctica are in contrast to the well known general warming effect of increasing CO2. The work contributes to explain the non-warming of central Antarctica since 1957.”
So, if a deranged terrorist fires a single Glock 19 round into a crowd of two hundred people, then the important number to work with there is the average bullet energy per person, right? The bullet itself cannot cause damage to the person it hits. This is excellent news for law enforcement.
So the greenhouse effect of CO2 can be positive, zero, or negative…LOLOL!! These asshats.
Imagine using that argument for your next speeding ticket? Sure you might have clocked me a 75 but my average was 60 so when you round them down I wasn’t speeding at all was I? Morons.
The question, then, is, “Does an average flux give us anything to work with as a cause of anything real in the actual world?”
You can calculate it. You can say that it exists mathematically. But can you apply it anywhere that is rational for a practical effect? No — it is a fantasy average that has no physical effect in any world that we live in.
The average energy per person of a Glock 19 bullet that kills one person within a two-hundred-person crowd is a ridiculous mathematical figure. The figure worth noting is the energy of the bullet that killed the one person. Talk to the relatives of the person killed — comfort them about the average bullet energy per person that could not, by itself, have killed their son — that the terrorist who fired that Glock round is not totally to blame for their son’s murder. Yeah, real comforting, isn’t it ?
Here’s the pdf in case you want to muse over these things nothing like reading a pdf on the pot. The author is a Dr. Holger Schmithüsen, he’s a Polish Meteorologist with a PhD whose thesis was on the greenhouse effect in the Antarctic. So apparently just use the word “greenhouse effect” in a thesis and you pretty much get a PhD. Someone should do an experiment in this write the most ridiculous paper and try and get it published in a “peer reviewed” journal. Make sure to use keywords like CO2 increase temperature, global warming crisis, and greenhouse effect global repercussions of fossil fuels and you’ll be praised for your work. His paper must be part of the ‘thousands of peer reviewed evidence of global warming’.
You know for a while I couldn’t figure out this ‘hundreds of papers show CO2 doesn’t increase plant growth’ so I looked into it myself. As it turns out thousands of experiments have proven CO2 increase growth ‘in mostly all plants’. So what they did is take the very few plants that it didn’t effect and produced hundreds of experiments to show it doesn’t increase growth.
I’ve also heard many times that some of these CO2 increases shown from Hawaii were secretly done near active volcanoes and that’s why I don’t trust this Co2 increase. Experiments have been done that shows CO2 is not a constant in the atmosphere and is not a well mixed gas and it changes seasonally, decreasing after cold weather and increasing after warmer weather.
Click to access BzPM_0692_2015.pdf
Click to access Schmithuesen.pdf
Antartica does not warm up because the rest of the world does not either. It has not for the last 20 or so years, CO2 or not. Roy Spencer proved it. Is there any higher God then him on that matter to contradict him ? What a bunch of 2 faces morons !
For those interested, I have destroyed the Green Plate Experiment shown here:
https://app.box.com/s/5wxidf87li5bo588q2xhcfxhtfy52oba
I give explanation and math here:
http://www.drroyspencer.com/2019/06/a-simple-no-greenhouse-effect-model-of-day-night-temperatures-at-different-latitudes/
Search link for
“sneaky fraud”
and read all MY comments below it.
Boreholes,
For those who might find it interesting and then can explain it better to me if so inclined 🙂
http://joannenova.com.au/2012/11/the-message-from-boreholes/
Ask astronauts about temperatures in and out of the sun…..
Is It Hot Or Cold At The International Space Station?
ISS (International Space Station)The ISS is a manned space station that continuously revolves around the Earth (Photo Credit : Wikipedia
What’s the average internal temperature of the ISS?
Short answer: As you can imagine, the temperature on the ISS is neither too high nor too low; rather, it’s maintained at a comfortable level as a result of certain elaborate processes and equipment so that the crew and the technical systems onboard continue to function without any problems.
The ISS experiences a wide range of temperatures
Orbiting the planet at an altitude of 330-435 kilometers above the surface of Earth, the International Space Station experiences a wide range of temperatures. Since it continuously revolves around the planet, sometimes it’s on the sunlit side of the Earth, while at other times, it’s on the dark side.
When the ISS faces the sun, the (external) temperature it experiences is around 250 degrees Fahrenheit (121 Degrees Celsius). On the other hand, when it’s on the side when our planet completely blocks out the sun, the thermometers plummet to minus 250 degrees Fahrenheit (-157 degrees Celsius).
Hmm, me thinks Joe is onto something.
What happened to this baby ?
The ice sheet may be melting in the periphery at low altitudes, but not in the main land at high altitude.
I wonder why they did not update it ? Not their favorite ?
Sorry,
Red line above 1500 meters altitude. Blue line below 1500 meters
Is there an analogy to be made in the calculations of ‘lift’ with regard to an airplane and attempts to average a snapshot of the Sun’s energy intercepted by Earth over the entire surface area of Earth?
What value would there be in averaging the lift over the entire surface area of the plane?
Thomas H asked, “What value would there be in averaging the lift over the entire surface area of the plane?”
ANSWER: By figuring the average lift over the whole surface area of the plane, you could then posit a “green air effect”, in order to account for the fact that the plane remains more aloft than it otherwise would without the “green air effect”. The “green air effect”, thus, slows descent, thus creating greater lift. The alternate explanation would have to do with “back lift”. I’ll let you do the math. (^_^)
Do you guys get notifications when I “like” your comments? The phone WordPress app has a star to like comments…I use it a lot for you guy’s comments…but I wonder if it actually does anything…
Robert K. – Excellent response! You summed up my speculation nicely. I think the same premise applies to buoyancy calculations for boat/ship hulls. What value would there be in averaging a hull’s displacement over the surface area of the entire vessel?
Joseph, you responded to several of my comments years ago and that gave me the confidence to continue my efforts in steering some of the online dialog (WUWT) towards recognizing that Carbon is not pollution. You might recall such gems as:
If Carbon were pollution we wouldn’t want to consume it.
Yet there are carbonated beverages.
If pollution is harmful to Life, and all Life is Carbon Based.
Then ‘Carbon Pollution’ is saying that Carbon is harmful to Carbon Based Life Forms.
[Without implying my recent comments warrant a ‘Like’, just letting you know I have not received any notifications.]
Macha says:
2019/06/26 at 6:48 AM
“When the ISS faces the sun, the (external) temperature it experiences is around 250 degrees Fahrenheit (121 Degrees Celsius). On the other hand, when it’s on the side when our planet completely blocks out the sun, the thermometers plummet to minus 250 degrees Fahrenheit (-157 degrees Celsius).
Hmm, me thinks Joe is onto something.”
Where did this quote come from because I never believe the “the thermometers plummet to minus 250 degrees Fahrenheit (-157 degrees Celsius).” nonsense !
Minus157 degrees Celsius is ~116 K and it takes 3 weeks for the lunar surfaces to cool to this level and rocky surfaces never approach this level except in possibly deep craters at the poles that never receive the solar radiation directly.
Here is a graph produced from the Diviner data :-
As you can easily see the temperature of the rocky lunar surfaces drop from ~370 K to ~270 K in approximately ONE WEEK ! (Because 6 lunar hours = 6 x 29.5 Earth Hours = 7.375 Earth Days)
Then in total darkness for another fortnight the temperature drops by another ~60 K down to about 210 K.
Just how long does the ISS spend in Earth’s shadow ?
Besides the problem with Skylab’s heat shield, damaged during launch by a piece of the launch tower, demonstrated what happens with a partially damaged heat shield – Skylab became uninhabitable with internal temperatures approaching 70°C until they fitted the parasol solution to cool it down enough over several orbits so they could repair the damage. Essentially they stuck a big umbrella out to reflect the solar radiation.
Anyone who tells you near Earth orbit space is cold or that the CMBR has any relevance is stupid beyond belief.
149 million kilometres away is a huge radiator continuously pumping 62.94 million joules per second per square metre of its surface and at Earth’s orbit this reduces to ~1368 joules per second per square metre of any surface irradiated by this radiation.
The Earth’s “shadow” is tiny with respect to the volume of space where this occurs – look at images of Venus transiting the Sun – it is negligible.
Anyone who states near Earth orbit space is cold is an idiot of unimaginable proportions ! (Not implying any smear to Macha)
Sorry – I didn’t check the size of the Diviner graph.
For those interested, I made a Google Spreadsheet for calculating Solar Insolation by Latitude.
https://bit.ly/2NkwOWA
If you download it to your own Google Drive or Computer, there will be pretty colors 🙂
Enjoy. -Zoe
JP, you filter links?
Insolation by Latitude:
bit (dot) ly (slash) 2NkwOWA
JP
Finally had a chance to watch this, good stuff.
To all those folks trying to show Roy Spencer et al, the idiocy of their ways good luck. I tried over a decade ago (iirc) along with others during the infamous Yes Virginia cold does heat hot nonsense. But the piled highly deep Doctor rebuffed all claims that the laws of thermodynamics provide no such behavior. Not to mention Joe’s many discussions with these flat earth troglodytes.
The other item of note in all this is their gross lack of understanding of power. Atmospheric CO2 emissions have no usable power they will heat the few molecules at lower energy around and above them and they will all radiate to space as the energy flows down the atmospheric gradient. They (CO2) are freezing cold at altitude (mean free path ) and radiate a very narrow spectrum. The entire GHG concept is shite on it’s face. Hence my favorite rebuttal to this trash science of “how many ice cubes do you need to cook a steak?”
According to these assholes, if we inject -100C CO2 into our refrigerator at 2C the fridge should warm up “cuz muh GHG” .Yeah sure ok retards.
I think at this point Roy and the rest of these frauds are so over-committed to the entire sham, they will never cop to their egregious errors.
Honestly talking to these climate clowns reminds me of the brat on the playground who was always a sore loser and could always get a group of kids to lie and say he won when they all know he lost.
There is definitely some truth in that. Spoiled brats become bullies.
I might add that bullies become tyrannical murdering psychopaths as well.
I went to Tahiti in winter once (February). I did some snorkelling. The air was 28 °C and the ocean surface water was 28 °C. Talk about a coincidence 🙂
I did some calculations. The Earth is tilted 23,5 degrees but there is always a surface from -23,5 degrees to +23,5 degrees East-West by -23,5 degrees to +23,5 degrees North-South that is always directly exposed to the Sun. That surface receives an average of 870 W/m^2 (by double Integral). Subtracting an albedo of 31% leaves us with 600 W/m2. Since that energy is calculated as falling on a flat surface we have to reduce it to a round surface which for all practical purposes is almost flat in that region. I don’t care to do the exact math, so let’s say we loose 15%. This leaves us with 510 W/m^2 which gives us an average temperature of 35 °C. That’s a lot of energy. Now, most of the Earth’s surface is ocean in that region (all around the globe) and it take about 3136 times more energy to heat the same volume of water then of air 1°C, and I’m leaving out the latent energy of vaporisation. So, on the dark side of the Earth, when the cooling down occurs, 1 molecule of water cooling 1°C, can heat up about 3136 molecules of air 1°C.
It looks to me that the clouds do not slow the atmosphere from loosing it’s energy to space (Greenhouse effect). It looks to me that as the air in the atmosphere cools down, the moisture in that atmosphere and the hot water in the oceans heats it up (3136 to 1) and keeps it warm enough until the next day comes.
Do I get that right or do I miss something ?
Yah it’s just thermal energy retention.
This winter, I have big outdoor grilling plans, where I will load my grill with ice cubes, place a nice juicy steak on top, and wait to see how long it takes to cook on one side. Once I determine this, I will then know how long to cook the other side, as well as the total cook time for a steak over ice cubes.
It probably will take a little longer than conventional methods, but a 97% consensus of REAL climate scientists seem to support this revolutionary approach. How can I go wrong?
Pierre – I get an average for -23.5° to 23.5° of ~0.9722 and therefore an average flux of 1368 x 0.9722 = 1330 W/m2 and an albedo adjusted average of 917.7 W/m2.
Are you using the mean value theorem for integrals ? You should be.
Sorry I stuffed up the dimensions again – need to think before pressing enter.
Pierre – I get an average for -23.5° to 23.5° of ~0.9722 and therefore an average flux of 1368 x 0.9722 = 1330 W/m2 and an albedo adjusted average of 917.7 W/m2.
Are you using the mean value theorem for integrals ? You should be.
Rosco,
For the whole Earth’s surface, it’s easy, INT(x, -Pi/2, +Pi/2) INT(y, -Pi/2, +Pi/2) COS(x) * COS(y) dy dx which gives me 1368 * 4. Since this gives me the flux density average on a flat disk under the sphere, just divide by 2/Pi to get the half sphere surface dendity. I did the same for the surface -23,5 to +23.5 with gives me 1368 * 0.7975 * 0.7975 but the flat surface under is not a flat disk. It’s more like a hot dog. 2 parallel lines with rounded ends.
I’m getting too old for that kind of math 🙂 and anyway all I wanted was an approximation since the subject was more about water releasing more energy then air when it cools. thus warming the air..1 cubic meter of salt water at the surface in a hot ocean cooling down 1C can heat up more then 3000 cubic meters of air 1C. Stack them all up 1 on top of the other and you get a colunm more then 3000 meters high. Since the atmosphere is not at constant pressure, you can safely say that 1 square meter 1 meter deep of hot water cooling 1C can heat up 1 square meter of the whole atmosphere all the way up to the top, or just about, Why are there palm trees in Scotland so far north ?
Regards,
Rosco,
There’s a difference between an average by latitude, and an average by AREA.
There is more area between 10N to 0N and 90N to 80N.
Take a look at https://bit.ly/2NkwOWA
So both Pierre and Zoe agree that area is the right way to calculate the average irradiation over the sphere of the Earth ? We’re back to the argument that the average insolation calculated by area as 1368/4 is right ?
It is obvious there is a difference between an average by latitude, and an average by AREA.
So let me ask you a question tempered by the fact I haven’t worked through your spread sheet.
At -35° latitude on the 28th day of the year the value in cell I38 is 292. I presume this is W/m2 and represents the solar insolation on this day at this latitude – or am I mistaken ?
This is the fortnight when the Australian Open Tennis is played in Melbourne at nearly 38°S.
Melbourne during this period regularly has temperatures in the 40°C + range and I fail to see how 292 W/m2 is capable of creating this temperature.
Where I live at 27°C South I calculate a value of ~368.5 W/m2 from the spreadsheet values of 380 at 25°S and 357 at 30°S as a rough guesstimate. I have measured the temperature in my roof cavity in excess of 60°C and the temperature of a light tan coloured concrete path at >55°C during the late morning. I don’t see how 368.4 W/m2 can create these temperatures.
At day 98 – the 8th of April I presume – the spreadsheet has a value of 540 for the North Pole but an average value of 236 for 27°S.
I’ll bet you anything you like the North Pole never reaches the temperatures we experience on the 8th of April – we regularly run our air conditioner during the day at this time of year.
We have an average annual temperature of 22°C – with a range from over 40°C to winter minimums rarely reaching close to 0°C at night, mostly low teens. We regularly have daily maximums over 30°C from October to April.
Or am I simply completely misunderstanding your spreadsheet ?
Pierre
I confess I do not know what your double integral calculates.
I assume your expression “COS(x) * COS(y) dy dx” is the same as this from Wolfram Alpha:-
https://www.wolframalpha.com/input/?i=integrate+cosx*cosy+dydx
If I am mistaken please explain what I have misinterpreted.
OK – I plugged in the definite integral as shown here :-
What does 4 mean ??
How have you calculated the flux normal to the spherical surface ?
The cos factor for the latitude seems to be missing from your cartesian co-ordinates.
Again, if I am mistaken please explain what I have misinterpreted. It is 2 decades since I attended my University Engineering course so my maths is rusty and my brain somewhat less agile. (I never practised as an Engineer – I was an Environmental Health Officer).
Joseph you need a comment board for those who need one aside from these passionate discussions beyond my math skills. Well you don’t need one just would be nice if you had one I guess. I wish you had one for stupid comments because I come across them all the time. For instance this one:
“Ice ages are fiction – North America was never covered by a giant glacier – the water erosion on North America is clearly explained in dozens of independent historic records – same with the water erosion on Latin America, Australia, England, etc. – continental drift and pangaea are fictions like most of the idiotic “geological history” of Earth – it’s wild imaginations on steroids and propped up by cartoons – the fake experts on parade got it all wrong. Dozens and dozens of witnesses documented the cause of the water erosion across North America, Latin America, Australia, etc. – throughout the 1400’s, 1500’s, 1600’s and 1700’s – that water erosion is not from melting ice and it is not from “thousands of years ago” – but you are right about the heat and we are not being told the true numbers of people dying from these heatwaves across Africa, Arabia, India, etc. – more deaths ahead as the Antarctic ice shelves continue to rapidly melt and collapse as thousands of Pennsylvania and Massachusetts-sized glaciers slide into the sea … what comes next is physics.”
I love the fact he ended his rant on physics ha ha where do these people learn these things from?
We share the world with psychotics who purposefully salt the fields. Like Zoe said 50% of people just seem to enjoy making others stupid.
Rosco,
Look at your figure 7. If the Sun’s flux diminishes with lattitude as a function of a cosine then it must diminishe the same way in the longitudes. COS(x) for lattitude correction, COS(y) for longitude correction.. Therefore, for any point x,y on the sphere exposed to the sun, the flux is 1368 * cos(x) * cos(y)..The integral of COS is SIN. SIN evaluated from -Pi/2 to PI/2 is 2. 2* 2 is 4. 1368 * 4 gives us the mean flux on the imaginary disk behind the half sphere. A half sphere has a surface 2Pi greater then a disk. So, 1368 * 4 / 2Pi = 1368 * 2/ Pi is the mean flux on the half sphere.
Rosco,
“We’re back to the argument that the average insolation calculated by area as 1368/4 is right ?”
Right. But we both agree this is irrelevant to global temperatures. The correct way is 3D, not 2D surface fluxes.
“At -35° latitude on the 28th day of the year the value in cell I38 is 292. I presume this is W/m2 and represents the solar insolation on this day at this latitude – or am I mistaken ?
This is the fortnight when the Australian Open Tennis is played in Melbourne at nearly 38°S.
Melbourne during this period regularly has temperatures in the 40°C + range and I fail to see how 292 W/m2 is capable of creating this temperature.”
The length of day is 10.8 hours on January 28 in Melbourne.
292 * 24/10.8 = 659 W/m^2 => 54C
OUPPPPSSSSSS !!!
Just caught a giant one here ! A disk has a surface of Pi * r^2 whereas half a sphere has a surface of 2 * Pi * r^2. Therefore half a sphere has a surface twice the surface of a disk. The results should therefore be 1368 * 2 and not 1368 * 2 / Pi. But the average can’t be higher then the highest value. Where did I go wrong ? Just can’t see it ? HELP !
Not sure. You tell me. When I do a double integral I take a density (W/m^2) and sum it up over the whole surface which should give Watts. I have to divide that total by the surface to come back to W/m^2. The surface is a disk and no matter r, the total should be divided by Pi. So, the integral should be done on [ Cos(x) * Cos(y) ] / Pi. The answer will be 1368 * 4 / Pi. Since half a sphere is double the surface of a disk, 1368 * 2 / Pi. Same final answer but in the right order this time.
Did I get it right this time ? Hope so ! You tell me.
Hahhh !!!!!!!!!!!!!!! Old age !!!
OK – I clearly do not see what your spreadsheet is doing. I’m not disputing your skills in formulating this and it will take some time to figure out and I’m not meaning to nitpick BUT :-
The length of day in Melbourne on January 28 is actually 14 hours and 6 minutes so 292 x 24/14 = ~501 W/m2 => 33.4°C so some discrepency there.
And at 70°N on day 98 you have 507. If I use the value of 15.36 hour day length for Barrow from time and date.com the calculation you cited becomes 507 x 24/15.36 = 792.19 W/m2 = 70.65°C ?
Am I wrong again ?
I just stumbled upon a gem from NASA, how long before it gets caught and removed? Notice the discalimer above it, however, great stuff it’s like the Wiki Global Warming link under Joseph’s YT video.
“The Sun provides the energy that drives Earth’s climate…” Oops who kept this up?
https://earthobservatory.nasa.gov/features/Aerosols/page3.php
Ok, I think I finally figured it out. 1368 * COS(x) * COS(y) double integrated from -Pi/2 to Pi/2 gives us the total flux density. To get the mean flux density you have to divide the result by the surface on which the flux falls. For a half sphere it’s 2Pi. Therefore 1368 * 4 / 2Pi = 1368 * 2 / Pi = 871 W/m^2.
For the segment of the Earth from -23.5 to + 23.5 N-S and -23.5 to +23.5 E-W from the incident Sun, 1368 * COS(x) * COS(y) double integrated from -23.5 to + 23.5 N-S and -23.5 to + 23.5 E-W gives us the total flux density of 1368 * 0.7975 * 0.7975 = 1368 * 0.636 = 870 W/m^2. The surface of the partial sphere is the double integral of COS(x) dx dy which gives us 0.7975 * 0.26111 * Pi. This gives us an average fllux of 870 / 0.2082 / Pi = 1330 W/m^2. HOT !!!
Hope I got it right this time 🙂
Rosco… Looks like your 1330 is right on !
Rosco,
Thank you for checking. I found the problem: Days were skewed to Vernal Equinox, not Jan 1. I fixed it. Check it out now.
Pierre,
You are comparing a line to a semicircle.
| (
The ratio here is 2 to PI, assuming R = 1.
The problem involves a disc to a hemisphere, not a line to a semicircle. You’re missing a dimension.
-Zoe
Zoe,
No I’m not.
Total Flux density over the surface = 1368 * 4
Surface under the total Flux density = 2 * Pi (surface of a hemisphere)
Mean flux density = 1368 * 4 / 2Pi = 1368 * 2 / Pi = 871.
Same resonning for a partial sphere. Rosco did the math using a discrete method. I come to the same results using integrals. 2 people using 2 different methods getting same result got it wrong ?
Regards,
Pierre,
“Total Flux density over the surface = 1368 * 4”
??? Why TIMES 4?
The sun is a flat disc with radius equal to earth, having a uniform output of 1361 W/m^2 through out the disc. This is the solar constant standard.
The area of this disc is PI*R^2
The surface area of Earth’s hemisphere is 2*PI*R^2.
The ratio is 1/2.
In your math you only consider hemisphere area when R=1, so you get 2*PI.
There is no consideration for the sun disc.
But you need to transform a disc into a hemisphere, not just find hemisphere area when R=1.
Chris Marshall,
I see you changed your youtube avatar to you holding a lovely British Blue cat.
I have the same breed of cat. His name is Æthelstan, Stan for short.
They are so adorable.
Zoe,
His name was Bleu (but I called him Bujjy) he was my best bud for 16 years and a whopper of a cat his peak weight was 16.2 pounds. His brother, Grey outlived him by 2 years.
Now I have Penelope who imprinted on me as a kitten and follows me everywhere.We have 5 cats and are fostering 3 more until they get homes.
Blues are extremely affectionate cats. Bleu had an impeccable sense of timing too, he knew when it was dinner time every time. At night at exactly 10 he would put his paw on the dinner door and look at us, my wife took a video of it. When we got them as kittens they had blue and gray collars and looked exactly the same so we called them “blue” and “gray” to tell them apart. The names just sprung from there.
Cheers
and d he was my best bud for 16 years.
Zoe,
Look at rosco’s figure 7, The total flux density is given by the double integral of 1368 * COS(x) * COS(y) dy dx over the given interval for the complete hemisphere and is 1368 * 4 no matter the value of r. r is a constant and does not change anything just like 1368 does not. If I integrate COS(x) * COS(y) dy dx I get 4. If I integrate 1368 * r^2 * COS(x) * COS(y) dy dx I get 1368 * r^2 * 4 since r is a constant.. If I leave r out in the original integral, I leave it out of the surface integral. If i put it in I end up with f(r) in the numerator and in the denominator and they cancel out. A bigger r will give me a bigger total flux and at the same time give me a proportional bigger hemisphere to spread that flux on. They cancel out.
The way I understand it is :-
1. The Earth always presents a hemisphere to the Sun – this is continuous regardless of day-night or seasonal considerations. Half of the Earth is always irradiated by the Sun. The relative position of the Sun in the sky locally changes with season and time but Half of the Earth is always irradiated by the Sun.
2. Every unit normal vector to any point on a sphere always passes through the centre of the sphere. I use the term unit because it allows for simple multiplication of the incident TOA power and works for any radius.
3. The Sun’s radiation is considered to be plane parallel.
4. Therefore the unit normal vector at any point at any point on a sphere is determined by 2 factors – the angle of incidence and the position on the hemisphere.
5. As the Sun’s radiation is parallel at all latitudes and longitudes the only consideration is the angle of the unit normal vector to the tangent at that point.
6. Using standard trigonometry in 2 dimensions allows for the construction of something like figure 7 that has been quoted and the average of 0.6366 or 2/pi.
7. Rotating that semicircle doesn’t alter the fact that the unit normal vector is determined by the angle from the centre to the tangent at that point.
Consider the reverse of these unit vectors – I say reverse because these are shown coming from the sphere but they represent a 3D view.
This is all just an approximation to show that climate science’s belief that the Sun’s insolation on Earth can be modeled by using one quarter power is not only ridiculous but also they appear to be so mathematically challenged that they can’t even use the appropriate averaging factor.
The fact that solar panels are rated against 1 kW/m2 input shows how absurdly removed from reality climate science is anyway – 342 W/m2 on average would render solar panels as a complete waste of time but I guess they’d work partially at night.
Rosco & Pierre,
“standard trigonometry in 2 dimensions”
Need to use 3 dimensions. 2/PI can only come from 2-dimensional analysis or b
neglecting the sun disc.
As far as solar panels, there’s a good rule of thumb:
A m^2 solar cell in US Southwest will collect in 24hrs the same energy as burning 1 kg of typical BROWN coal.
I worked this out while on wall street.
I flip flopped renewables and fossil fuels as a full time job for several years. I was right ~84% of the time – which is considered excellent in the speculation industry.
Of course burning that coal will give you that energy in the here and now vs. 24hrs for the solar cell.
-Zoe
No matter where you are have a happy 4th of July, grill a burger or frank just for the CO2 if nothing else.
Rosco,
As far as I’m conserned, you are right on the button in your explanation !
Zoe,
The Sun is so far away and so big compared to the earth that it can mathematically be approximated as being flat and the rays comming from it and intercepted by Earth can mathematically be considered as parallel. Therefore only 2 dimentions.
In other words, the rays comming out of the Sun just a bit in the north, south, east or west direction fron Earth wont hit the Earth. Only the ones comming straignt from the center of the Sun facing directly the Earth. So you can mathematcaly consider the Sun as a point in space irradiating parralel rays..
Regards and a very good 4th of July (from a Canadian)
How about this ?
There are 2pi steradians in a hemisphere. With unit plane parallel irradiance the average irradiance over a hemisphere is 1/2Pi.
Rosco,
“With unit plane parallel irradiance the average irradiance over a hemisphere is 1/2Pi.”
Who said solar disc was a unit plane?
The area of solar disc is PI*R^2
The area of hemisphere is 2*PI*R^2
Ratio is 1/2
If the solar disc was a unit plane then you would be correct, but it’s not.
Zoe,
Wrong !
Look above at the picture the way the Earth irradiates it’s outgoing infrared energy. It’s in all directions. North, South, East, West, Everywhere. The Sun emits it’s energy the same way , in all directions. ONLY the ones emmited directly towards the Earth reach the Earth. All the others go to outer space to be lost and never hit the Earth. So the Sun although very big can be considered a point in space since only the energy comming from that point reaches the Earth. The energy coming from that so small a point so far away are parralel and are 1368 W/m2.
JP
You’ve been oddly diplomatic in this conversation. A new look for you to be sure. I was expecting more input from someone referred as “… a first class physicist…” -John Salamito comment on “On the Flat Earth Rants of Joe Postma” on PSI page.
Anyway, yeah you’ve been quiet.
No-one other than you has said anything even remotely like “Who said solar disc was a unit plane?”
Obviously the steradian interpretation is wrong because a steradian describes an area equal to radius squared on the surface of a sphere.
If you read correctly the only claim ever made is that given the distance the Earth is from the Sun the solar radiation is considered to be approximately plane parallel. This is not strictly true but the divergence is small and is considered negligible for the purposes of generalisations such as this discussion.
The divergence from plane parallel for the vectors from the Sun to the Earth is of the order of 0.0012°.
Hence the use of the graphic quoted previously as figure 7 where parallel vectors representing the solar radiation hitting the 2D representation of the hemisphere of the irradiated Earth. Note further that these vectors are also close enough to parallel not only from north to south but also from east to west.
There is absolutely no doubt that a vector incident on any point on a hemisphere can be parameterised into components parallel to the tangent line or plane and normal to the plane.
The normal component is the cosine of the angle subtended from the centre to the incident vector at the interception point and every one of these normal vectors passes through the centre of the circle. The tangential component is the sine of the angle. This is standard trigonometry used in statics and is used everywhere in science and engineering.
I borrowed this illustration from Alan Siddon’s article referenced previously – I hope he will not object.
It shows a valid transformation from 2D to 3D – not a standard integral but a plane rotation of a hemisphere about the centre keeping the incident vectors parallel and normal to the “equator” of the heimsphere.
The average unit irradiance over every one of these 2D hemispheres is 2/pi. The sum of all of them necessary to render a 3D surface is “n” times each semicircular plane and the average irradiance over the sum is n x 2/pi x 1/n = 2/pi.
This is good enough for me.
CM: I’m stalling on whether I should record the next video I feel like doing.
Is that what someone said about me? Very nice.
And no one has shown the correct solution to the average hemispheric projection. Sorry Rosco but that solution only works on a cylinder…on a sphere, you see that those geodesics have greater and greater space between them going away from the pole…on a cylinder that doesn’t happen. I’ll show you guys the correct solution soon…it’s a nice integral.
JP
You are brilliant which is why you should pursue a doctorate at some point, IF Trump gets re-elected opportunities for space exploration are going to sky rocket and the demand for top physicists will too. NASA might actually get back to doing real science for a change like the Orion Program. Sadly in most cases a doctorate is just a badge of passage but it does prove you are serious and capable of accepting larger responsibilities. I can’t imagine someone hasn’t approached you about furthering your career by now or at least suggested it?
Sometimes you just got to make things happen regardless of obstacles.I do understand you have a personal life as well and that does hold sway on choices.
Joe …it’s a nice integral.
Don’t tell me it has anything to do with sperical trigonometry… I’m too old for that 🙂
As we go further N or S the flux gets weeker and weeker but the area on which it is spread is smaller and smaller…
https://vincematsko.files.wordpress.com/2017/03/fig31.png?w=300&h=368
I have not figured it all out yet but why do I get the funny feeling the answer will be somewhere closer to 1368 * 4 then 1368 * 2 / Pi ?!?!
The integral is more intuitive and simpler if you exploit the existing symmetry.
Turning things around and around and around OH YA and the only thing I can figure out is this…
The double integral of 1368*COS(x)*COS(y) dy dx gives me 1368 * 4, but it’s the energy intercepted by a square 2r by 2r = 4r^2. So I have to reduce that energy to the energy included in the included disk, then reduce that to a hemisphere…
1368 * 4 * Pi / 4 / 2Pi = 1368 / 2 = 684 W/m2
Can’t do any better… You tell me !!!
Thanks, Joe.
My wife and I had a right good laugh. I sort of like Roy (lukewarmer) but find some of his blogs subjects and comment replies a bit reliant on faith rather than science. His practice of driving about the streets, pointing his thermocouple at the sky measuring back radiation to “prove” it exists never fails to raise my eyebrows ( well, one at least in the best Roger Moore fashion). Lindzen and Monckton are much the same. Nice guys trying to rail against climate alarmism but at the same time falsely recognising the role of CO2 (albeit small) in the IPCC “global heating” (latest alarmist term) scenarios.
I have a university background in maths and physics (UK) but it was long before Pontius was a pilot so I had to re-educate myself on what i knew then to what I know now. But I do appreciate your long standing efforts to fight against climate sophistry ably assisted by Rosco over the years.
While I’m here, can you explain to me the mantra of both climate alarmists and climate lukewarmers that proclaims, “Scientists say that doubling pre-industrial carbon dioxide levels will likely cause global average surface temperature to rise between 1.5° and 4.5° Celsius “. Where does this assertion come from?
I keep looking at your pig on a spit visual representation and seeing Al Gore there. Am I a bad person?
leitmotif,
I’m the person who did that pig-on-a-spit visual for Joe, and I regret not using Al Gore, now that you mention it.
What a lost opportunity!
Robert Kernodle.
A porcus moment. You are a bad person too. 😀
AL Gore on that spit with Trump turning it would induce climate rage all across the land especially if you add the IPCC reports as kindle fire.
The best I have been able to come up with is to use the rotation of a semicircle in the x-y plane about the x-axis. This produces a sphere.
Viewing this transformation from the x direction and assigning this as the solar radiation direction – from x -> the origin – one would view lots of half “protractors” rotating to form a hemispherical surface similar to the figure Alan Siddons uses.
The average on each of these is 2/pi.
I can’t see why this isn’t valid and can’t think straight anymore.
z = 1368 * cos(x) * cos(y) does not seem to be the right equation because for any x value there is a restraint on y in the form of y = sqrt (1 – x^2). Not all values of y are permitted.
I’m facing a wall !!!
Cheers @ leitmotif
“While I’m here, can you explain to me the mantra of both climate alarmists and climate lukewarmers that proclaims, “Scientists say that doubling pre-industrial carbon dioxide levels will likely cause global average surface temperature to rise between 1.5° and 4.5° Celsius “. Where does this assertion come from?”
It comes directly from their flat-Earth physics & fake greenhouse effect. It comes directly & exclusively & solely from their position that sunshine/solar input is only -18C, and therefore the climate creates and heats itself with this fake version of their greenhouse effect where “backradiation” from the atmosphere makes the atmosphere create its own temperature & the climate. All this comes from their peer-reviewed literature which the climate people happily produce and publish under this paradigm just described. It comes from their position that with additional CO2 in the atmosphere and in the climate, that the climate will now be able to create itself some more. That is…without any embellishment or misrepresentation, the absolute state of this science.
And, of course, all of that paradigm based upon *literal* flat Earth physics…from direct creation out of, or absolute equivalency to, flat Earth theory and how reality would have to work if you could have such things as flat Earths.
leirmotif,
The religion claims energy in = energy out. Yet they actually do surface flux density in = surface flux density out.
They claim that Earth receives and emits 238 Joules every Second, every m^2 (255°K).
But …
Heat required to raise temperature by 1°C:
Q = m * Cp * dT
For CO2 to change atmospheric temperature by 1°C:
m (air) * Cp (air) * 1 =
m (co2) * Cp (co2) * dT
Givens:
m (co2) / m (air) = 0.0006
Cp (air) = 1.01
Cp (co2) = 0.83
Our equations becomes:
1.01 = 0.0006 * 0.83 * dT
dT = 2028°K
CO2 needs 2028°K (959,080 W/m^2) worth of energy to raise atmo by 1°C. But it only has 255°K (238 W/m^2).
Where does it magically get the extra energy?
Rosco,
The math of 1/4 has been confirmed by satellites.
It’s good to point out mistakes in science, but sometimes one can go too far …
https://www.mdpi.com/2072-4292/11/7/796/pdf
“A combination of ERB scanning instruments leads to an EOR
value of 336.1 ± 4.7 W/m^2”
“The math of 1/4 has been confirmed by satellites.”
The whole point of these videos is that the math of 1/4 is nonsense and completely non physical !
Whether or not a hemisphere has an average irradiance of 2/pi by trigonometric considerations or 0.5 by area considerations is only a side issue – although it is a delicious irony to consider all of climate science is invalidated if the trigonometric argument is correct.
There is ZERO DOUBT the maximum radiative power incident on the Earth is not 1/4 of the solar constant !
Climate science is all based on the premise shown from http://people.atmos.ucla.edu/liou/Lecture/Lecture_3.pdf
Using the hypothesis of radiative equilibrium to make a back of the envelope calculation of Earth’s average emission temperature may be fair enough. But climate science expects us to believe that the average emissive power of 1/4 S(1-a) is actually the average INCIDENT radiative power incident over both night and daytime at the same time and this is nothing short of gobbledygook and only believers in witchcraft or fairy dust could support it.
Are you asserting that the satellites measuring the incident solar radiation DO NOT record ~1361 W/m^2 and actually record 336.1 ± 4.7 W/m^2 ?
Rosco,
“by trigonometric considerations or … by area considerations”
Both are trying to find area dilution from solar disc to earth sphere.
“DO NOT record ~1361 W/m^2 and actually record 336.1 ± 4.7 W/m^2 ?”
1361 is the solar disc.
340 is the earth sphere.
Are you trying to find an answer for sun providing 15°C? Is that what you’re doing? That would be great, but how do you get over 16,000 W/m^2 for Venus from the 2604 W/m^2 it receives?
While we’re on the subject of witchcraft or fairy dust:-
Climate science says – http://people.atmos.ucla.edu/liou/Lecture/Lecture_3.pdf – that Venus has a HUGE “greenhouse effect”.
It is so powerful that the surface of Venus irradiated by 132 W/m2 can radiate at 16,101 W/m2 – 730 Kelvin.
BUT if we consider Venus’ atmosphere at 96% CO2 is almost comparable to an ideal gas we can calculate a temperature using NASA data (presumably 100% reliable – they did put a man on the Moon after all) as I did on PSI in 2014.
The calculation is PV(unit volume 1 cubic metre chosen) = n R T.
92,000 (mb) x 1,000 (litre/cubic metre) = 65,000 (g/cubic metre)/(3.45 (g/mole) x 0.082 x T
T = 92,000/(0.082 x 65,000/43.45) = 749.98 Kelvin.
Greenhouse believers completely dismiss the concept that compression of an atmosphere by gravity can cause a temperature. The compression of gas in a car tyre for example causes only a temporary increase in temperature they say.
But compressing an atmosphere in an environment where conduction doesn’t exist is an entirely different concept.
Yet every atmosphere has a temperature gradient and the temperature increases as one descends through the atmosphere. NASA cite the internal temperature of Jupiter as possibly as high as 24,000°C – a remarkable greenhouse effect from 50.5 W/m2 irradiance and almost no “greenhouse gases”.
I used NASA’s Planet Fact Sheets to do similar calculations for all of the planetary atmospheres and found the calculations provided answers in accordance with the data. Earth is different as the atmosphere is not comparable to an ideal gas but the answer fits within the top of the diurnal range cited by NASA and Mars agrees with data from the Viking Lander site.
Of course none of this is completely “right” but it is interesting and far more logical than spreading the irradiation over the whole of a sphere by dividing insolation by 4 !
Rosco,
I’m not disputing the fact that the greenhouse effect is junk science, only your calculation for global average insulation.
Are you claiming Venus’ global average insolation is not 661 (2604/4)?
Zoe
“1361 is the solar disc.
340 is the earth sphere.”
So you assert the real time insolation at any point on Earth’s surface at any time is S/4(1 – a) ?
There is no solar disk – theSun is a “sphere” at a disatance of an average of 149.9 million kilometres emitting ~62.94 million joules per square metre per second into 3 D space over its entire surface area of ~4.Pi.695,700^2 at an effective emission temperature of ~5772 Kelvin.
@ leitmotif
Thanks for the compliment.
Cheers
Thanks for the compliment leitmotif.
“I’m not disputing the fact that the greenhouse effect is junk science, only your calculation for global average insulation.
Are you claiming Venus’ global average insolation is not 661 (2604/4)?”
I don’t care about the average global insolation !
Venus’ global Average insolation is most decidedly not 661 (2604/4) ! So you assert that 661 x 0.2 = 132 W/m2 can result in surface emissions of 16,101 W/m2 ?
The satellite document does NOT measure incoming solar radiation – they simply divide the TSI by 4 !!
“The Earth energy imbalance mentioned above may be calculated from the extant space-based TSI and EOR measurements listed above, EEI = TSI/4 − EOR, leading to an EEI of 4.3 ± 4.9 W/m2 (note that TSI must be divided by 4 to compute the incoming solar flux arriving at the global TOA).”
This is complete nonsense – incoming solar radiation TOA is 1361 W/m2 or whatever the actual solar radiation is.
“Turning our attention to the example of Langley’s greenhouse experiment on Pike’s Peak in Colorado (mentioned by Arrhenius, 1906b), we may be tempted to ask how it is that a greenhouse can reach such high temperatures. Qualitatively, we may attribute the difference between the 15ºC mean surface temperature and the 113ºC observed in Langley’s greenhouse to the fact that noon-time radiation at the surface is three to four times as intense as the mean radiation over the whole of the earth’s surface.”
Using the SB equation 1361 W/m2 is the radiation associated with a temperature of ~394 K or ~121°C. Given that Langley recorded a temperature of 113°C it is impossible that 1361/4 can generate this temperature !
If you calculate the ratio of the radiating temperature for Venus cf to Earth then Venus should have temperatures ~1.176 times Earth’s.
At sea level Earth is ~15°C and Venus at ~1000 mb is 1.176 times = ~339 K and the temperature profile data for a Venus probe confirms that at similar atmospheric pressures the ratio of 1.176 holds pretty well.
1.176 is the square root of 149.6/1082.1.
The most obvious stupidity of using the 1/4 ratio is it ignores time.
Venus has a day length of almost 4 Earth months. Plenty of time for a surface to be roasted but still not enough to explain the surface temperature unless you factor in compression of an atmosphere.
The Moon has a night time exceeding a fortnight on Earth but during this time the already cool surfaces radiate at a verily slow rate of temperature decrease. From noon to 6pm lunar time the rocky surfaces on the Moon’s equator’s rate of temperature decrease is of the order of 0.7°C per Earth hour.
Time is obviously an important consideration but they ignore it entirely and devise all sorts of ludicrous explanations to defend a hypothesis that has zero empirical evidentiary support.
Rosco,
What are you arguing?
You believe the global average solar flux is 389 W/m^2 (15°C)?
I’m turning my head around and around and around, I just can’t see where I’m wrong. 1368 * 4 is the total flux density but distributed on a square 2r * 2r = 4r^2. Just redistribute that total on the corresponding hemisphere’s surface of 2Pi * r^2 to get the average…
1368 * 4 / 2Pi = 1368 * 2 / Pi = 871.
I have that nasty feeling that 1368 / 2 = 684 might be right but don’t know why.
Joe, I feel like a fish at the end of a line. Please reel me in if it’s not right !
Pierre I’ve got a real nice post coming up ASAP for you. I can see you have good mathematical literacy, so once you see the solution I am sure you will be delighted in the symmetry and theorems which are used.
Zoe I’m not arguing that the global average insolation is any particular value.
I am saying that there is no physically real value to an average global insolation and it is most certainly not 1/4 of the TOA radiation – because it doesn’t exist !
The only physically meaningful average is the radiance over a hemisphere because this is where it actually exists. And even if it is 0.5 or 2/pi it really is not meaningful either because it does not relate to any real temperature except that it would be really nice if climate science was completely wrong.
Rosco, Lord Kelvin summed it up, nicely. In fact it might have been the 20th century AGW hypothesis he was future-projecting his thoughts onto all those years ago. Clever guys those Belfast boys.
“I often say that when you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind; it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science, whatever the matter may be.”
Joseph,
I’m really hoping at some point you cover Venus and Mars and the CO2 conspiracy on a video. Be nice for someone to hear the science behind it for a change.
I told my wife i was going to lie out in the sun in Calahonda, Costa Del Sol, from 8am to 8pm in July.
She told me I would be burnt to a cinder with the midday heat.
I told her not to be ridiculous. The average temperature during that 12 hour period was only 68F.
What do women know, anyway? Heh? I’m a guy for whom the sun is always a quarter full, not three quarters empty.
Hear from and speak to you all when I get back with my even tan.
(Sounds of ambulances in the background)
CD Marshall, that’s easy.
Men are from Mars, Women are from Venus.
They both thrive on different levels of CO2 it seems.